Sum polynomial and exponential increments · source reference

Printed question

A sequence of real numbers $u_1,u_2,u_3,\ldots$ satisfies

$$u_n=u_{n-1}+n+2^n\quad(n\ge2).$$

Given $u_1=4$, show that $u_n=\frac{(n+a)(n+b)}2+2^{n+1}$, where $a,b$ are integers to be determined. Hence determine, with reasons, the behaviour of the sequence. [5]

Printed working

$$\begin{aligned}u_n&=u_{n-1}+n+2^n\\&=u_{n-2}+(n-1)+2^{n-1}+n+2^n\\&=u_{n-2}+(n-1)+n+2^{n-1}+2^n\\&=u_{n-3}+(n-2)+2^{n-2}+(n-1)+n+2^{n-1}+2^n\\&=u_{n-3}+(n-2)+(n-1)+n+2^{n-2}+2^{n-1}+2^n\\&=\cdots\\&=u_1+2+3+\cdots+(n-2)+(n-1)+n+2^2+2^3+\cdots+2^{n-2}+2^{n-1}+2^n\\&=4+2+3+\cdots+n+2^2+2^3+\cdots+2^n\\&=4+\frac{(n-1)(n+2)}2+\frac{2^2(2^{n-1}-1)}{2-1}\\&=\frac{(n-1)(n+2)}2+2^{n+1}.\end{aligned}$$

Hence $a=-1,b=2$ (or interchanged).

Since $u_n>u_{n-1}>\cdots>u_1$ for $n\ge2$, and as $n\to\infty$, the source additionally writes $\frac n2(n+1)\to\infty$ and $2^{n+1}\to\infty$ (a comparison polynomial; the exact formula above uses $\frac{(n-1)(n+2)}2$), $u_n\to\infty$: increasing and divergent.

Teacher notes

Use the recurrence formula to obtain $u_n$ in terms of $u_{n-1}$. Repeat to obtain $u_{n-1}$ in terms of $u_{n-2}$ and so on until a pattern can be observed; then generalize to $u_1$.

Describe behaviour with (1) increasing, decreasing or oscillating and (2) convergent (to a specified limit) or divergent.

Printed answer key

−1, 2; increasing and divergent

Mathematical review and qualifications

$u_n=4+\sum_{k=2}^n k+\sum_{k=2}^n2^k=4+[n(n+1)/2-1]+[2^{n+1}-4]=\frac{(n-1)(n+2)}2+2^{n+1}$. Thus $a=-1,b=2$ (or interchanged). Every increment $n+2^n$ is positive; the explicit expression tends to $+\infty$. The sequence is strictly increasing and divergent.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: RECURRENCE RELATIONS. [2015/AJC/Promo/11(b)] Question PDF pages 2; solution PDF pages 5. The original files remain private QA records outside this native content.